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Homework: conservation of momentum

  1. May 1, 2006 #1
    Hi,
    could you please help me with following:
    ----------
    two boats traveling in opposite direction with speed v1 and v2

    when they are passing by they interchange a parcel of same mass M=50kg

    as a result second boat stops and first boat continue travelling in same direction with speed u1=8.5m/s

    calculate v1 and v2 when you know masses (including M) m1=1000kg and m2=500kg

    ------------

    I wrote
    m1*v1+m2*v2=m1*u1

    but I am not able to find out second equation. I tried kinetic energy but
    I think it is not conserved.

    Thank you
    Michael
     
  2. jcsd
  3. May 1, 2006 #2

    Hootenanny

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    Welcome to PF,
    What do you mean when you say;

    ~H
     
  4. May 1, 2006 #3
    Sorry my English is very poor.

    one boat is going East and the other West, when they meet
    an object of mass 50kg is handed over from boat1 to boat2 and at the same time other object of same mass is handed from boat2 to boat1


    Thank you
    Michael
     
  5. May 1, 2006 #4

    Hootenanny

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    No problem. I would say for this question you have to assume that kinetic energy is conserved. You correctly set up you momentum equation, if you then setup a kinetic energy equation, you have two equations (one for kinetic energy and one for momentum) and two unknowns (v1 and v2), you can then solve both equations simultaneously to obtain v1 and v2.

    ~H
     
  6. May 1, 2006 #5
    thank you, but ...

    I wrote two equations

    v1*m1+v2*m2 = u1*m1

    and

    (m1*v1^2)/2 + (m2*v2^2)/2 = (m1*u1^2)/2

    the result is not M dependent ?
    when I solve the equations I get wrong answers.
    (correct is v1=9m/s and v2=-1m/s)

    Or when and why should M in kinetic energy equation be ?

    Thank you very much

    Michael
     
  7. May 1, 2006 #6

    Andrew Mason

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    This is not an elastic collision. Energy is not conserved. Only momentum is conserved. You know that the 50 kg mass travelling at v1 when caught by the second boat causes the second boat to lose all its momentum. Work out the formula for that. That should give you the ratio of the two speeds and will enable you to express v1 in terms of v2.

    Then work out the equation for conservaton of momentum for the other boat and substitute your value for v1 in terms of v2. That will give you the answer.

    AM
     
  8. May 1, 2006 #7

    Hootenanny

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    I've just worked through the calculations now (I've only just got a spare minute), and I made an incorrect assumtion that the collision was elastic (hence kinetic energy being conserved). If you consider the momentum of the package travelling at v1, you should be able to obtain a ratio of the two velcoities (one will be negative).

    ~H

    Edit: AM got there before me.
     
    Last edited: May 1, 2006
  9. May 1, 2006 #8

    Curious3141

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    Just a small niggling point (I'm sure you know, it's just that others might get the wrong idea) - energy (the sum total of all forms) is *always* conserved. Kinetic energy is not conserved (in this case). :smile:
     
  10. May 1, 2006 #9
    The two equations are

    M*v1 + (m2-M)*v2 = 0

    and

    (m1-M)*v1 + M*v2 = m1*u1 ?

    I hope so. At least they give correct answer, I use them.:smile:

    Thank you
    Michael
     
  11. May 1, 2006 #10

    Hootenanny

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    Yup, they look correct to me.

    ~H
     
  12. May 1, 2006 #11

    Andrew Mason

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    You could also use:

    [tex]m_1v_1 + m_2v_2 = m_1u_1[/tex]

    which expresses the conservation of momentum of the whole system.

    AM
     
    Last edited: May 1, 2006
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